The number of calls coming per minute into a hotels reservation center is Poisson random variable with mean 3. 
The number of calls coming per minute into a hotels reservation center is Poisson random variable with mean $3$.
  (a) Find the probability that no calls come in a given $1$ minute period.
  (b) Assume that the number of calls arriving in two different minutes are independent. Find the probability that at least two calls will arrive in a given two minute period.

Can someone help me out how to answer this? I just don't get the way on how to solve this.
For A) $P(x=0)$, I know that Poisson is $\frac{e^{-u}(x)^u}{x!}$ where $u$ is the mean. I think the answer is $0$? 
for B) I totally don't get it I just know that it is $P(X>2)$
 A: a) No, not quite, you got the pmf wrong. We have that the number of calls in a one minute interval $(0,t = 1]$ has a rate of 3. If we call that $X(t)$, then $X(1)\sim\text{Pois}(\mu t = 3\cdot 1)$. Hence
$$P(X(1) = 0) = e^{-\mu}\frac{\mu^x}{x!}= e^{-3}\frac{3^0}{0!} = e^{-3} $$
where $0! = 1$.
b) It asks $$P(X(2) \geq 2) = 1-P(X(2)<2)$$
where it is easier to use the complementary probability.
Note: Technically, a given minute does not have to be the interval $(0,1]$; it should be $(s, s+t]$ so that $X_{(s+t)}-X_{(s)}\sim\text{Pois}(\mu(s+t-s))$ where $t$ will equal one.
A: Let $X_t\stackrel{\mathsf{iid}}\sim\mathsf{Pois}(3)$ for $t=1,2,\ldots$. Then the probability sought in part (a) is $$\mathbb P(X_1=0) = e^{-3}\approx 0.050, $$
and the probability sought in part (b) is
\begin{align}
1 - \mathbb P(X_1+X_2<2) &= 1 - (\mathbb P(X_1+X_2=0)+\mathbb P(X_1+X_2=1))\\
&= 1 - e^{-6}(1+2\cdot3)\\
& = 1 - 7e^{-6}\approx 0.983.
\end{align}
For what it's worth, I happened to stumble upon this problem here: https://www.hackerrank.com/challenges/poisson-distribution-3
I submitted those values and the website marked it as correct.
